Second time derivative of function?

I have defined a model of displacement 'x' and represented its first derivative with 'v' using the code below. However, when I want to differentiate 'v' using the same function it does not work. Is there a different approach to differentiating derivatives in MATLAB? here is the code:

h = 60 ; 
b =130 ; 
r = 28 ; 
w = 2*pi; % angular velocity

t =0:(1/36):1 ; 
x = b*(r*sin(w*t))./(h-r*cos(w*t));
v = diff(x)./diff(t) ;   % derivative of x w.r.t t
a = diff(v)./diff(t) ;   % derivative of v w.r.t t

Answer 1

[Aquatris]

It does not work, because the length of diff(x) and diff(t) are not the same. diff function takes the difference between each element. However returns 1 element less than the original vector since there is no way to find the difference for the last element.

Since you have fixed step size you can use

v = diff(x)./(1/36); 
td = t(1:end-1); % time vector for plotting first derivative
a = diff(v)./(1/36) ;tdd = t(1:end-2)    
tdd = t(1:end-2); % time vector for plotting second derivative

Alternatively you might try symbolic toolbox to derive the derivative of the expression symbolicly and then plug in numbers.

[Christian Aranas]

Thank you so much for the context

Answer 2

[KSSV]

diff reduces the dimension of vector by one. You may use gradient

    h = 60 ; 
    b =130 ; 
    r = 28 ; 
    w = 2*pi; % angular velocity
    t =0:(1/36):1 ; 
    x = b*(r*sin(w*t))./(h-r*cos(w*t));
    v = gradient(x)./gradient(t) ;   % derivative of x w.r.t t
    a = gradient(v)./gradient(t) ;   % derivative of v w.r.t t

Or, reduce the time step by one and use _diff_

h = 60 ; 
b =130 ; 
r = 28 ; 
w = 2*pi; % angular velocity
t =0:(1/36):1 ; 
x = b*(r*sin(w*t))./(h-r*cos(w*t));
v = diff(x)./diff(t) ;   % derivative of x w.r.t t
a = diff(v)./diff(t(1:end-1)) ;   % derivative of v w.r.t t

[Beau Olafson]

That gradient code 'literally' saved my life. I was so dang lost until then!